/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Outermost Termination of the given OTRS could not be shown: (0) OTRS (1) Thiemann-SpecialC-Transformation [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) MRRProof [EQUIVALENT, 19 ms] (13) QDP (14) PisEmptyProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) TransformationProof [EQUIVALENT, 0 ms] (22) QDP (23) UsableRulesProof [EQUIVALENT, 0 ms] (24) QDP (25) QReductionProof [EQUIVALENT, 0 ms] (26) QDP (27) Trivial-Transformation [SOUND, 0 ms] (28) QTRS (29) AAECC Innermost [EQUIVALENT, 0 ms] (30) QTRS (31) DependencyPairsProof [EQUIVALENT, 0 ms] (32) QDP (33) DependencyGraphProof [EQUIVALENT, 0 ms] (34) AND (35) QDP (36) UsableRulesProof [EQUIVALENT, 0 ms] (37) QDP (38) QReductionProof [EQUIVALENT, 0 ms] (39) QDP (40) QDPSizeChangeProof [EQUIVALENT, 0 ms] (41) YES (42) QDP (43) UsableRulesProof [EQUIVALENT, 0 ms] (44) QDP (45) QReductionProof [EQUIVALENT, 0 ms] (46) QDP (47) TransformationProof [EQUIVALENT, 0 ms] (48) QDP (49) TransformationProof [EQUIVALENT, 0 ms] (50) QDP (51) TransformationProof [EQUIVALENT, 0 ms] (52) QDP (53) TransformationProof [EQUIVALENT, 0 ms] (54) QDP (55) TransformationProof [EQUIVALENT, 0 ms] (56) QDP (57) TransformationProof [EQUIVALENT, 0 ms] (58) QDP (59) TransformationProof [EQUIVALENT, 0 ms] (60) QDP (61) TransformationProof [EQUIVALENT, 0 ms] (62) QDP (63) MNOCProof [EQUIVALENT, 0 ms] (64) QDP (65) NonTerminationLoopProof [COMPLETE, 0 ms] (66) NO ---------------------------------------- (0) Obligation: Term rewrite system R: The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 Outermost Strategy. ---------------------------------------- (1) Thiemann-SpecialC-Transformation (EQUIVALENT) We applied the Thiemann-SpecialC transformation to transform the outermost TRS to an innermost TRS. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_f(result_f(x)) -> go_up(x) check_id(result_id(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) TOP(go_up(x)) -> REDUCE(x) REDUCE(f(x_1, x_2, x_3)) -> CHECK_F(redex_f(x_1, x_2, x_3)) REDUCE(f(x_1, x_2, x_3)) -> REDEX_F(x_1, x_2, x_3) REDUCE(id(x_1)) -> CHECK_ID(redex_id(x_1)) REDUCE(id(x_1)) -> REDEX_ID(x_1) CHECK_F(redex_f(x_1, x_2, x_3)) -> IN_F_1(reduce(x_1), x_2, x_3) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_1) CHECK_F(redex_f(x_1, x_2, x_3)) -> IN_F_2(x_1, reduce(x_2), x_3) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_2) CHECK_F(redex_f(x_1, x_2, x_3)) -> IN_F_3(x_1, x_2, reduce(x_3)) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_3) CHECK_ID(redex_id(x_1)) -> IN_ID_1(reduce(x_1)) CHECK_ID(redex_id(x_1)) -> REDUCE(x_1) REDUCE(s(x_1)) -> IN_S_1(reduce(x_1)) REDUCE(s(x_1)) -> REDUCE(x_1) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_f(result_f(x)) -> go_up(x) check_id(result_id(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_1) REDUCE(f(x_1, x_2, x_3)) -> CHECK_F(redex_f(x_1, x_2, x_3)) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_2) REDUCE(id(x_1)) -> CHECK_ID(redex_id(x_1)) CHECK_ID(redex_id(x_1)) -> REDUCE(x_1) REDUCE(s(x_1)) -> REDUCE(x_1) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_3) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_f(result_f(x)) -> go_up(x) check_id(result_id(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_1) REDUCE(f(x_1, x_2, x_3)) -> CHECK_F(redex_f(x_1, x_2, x_3)) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_2) REDUCE(id(x_1)) -> CHECK_ID(redex_id(x_1)) CHECK_ID(redex_id(x_1)) -> REDUCE(x_1) REDUCE(s(x_1)) -> REDUCE(x_1) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_3) The TRS R consists of the following rules: redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. top(go_up(x0)) reduce(f(x0, x1, x2)) reduce(id(x0)) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_1) REDUCE(f(x_1, x_2, x_3)) -> CHECK_F(redex_f(x_1, x_2, x_3)) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_2) REDUCE(id(x_1)) -> CHECK_ID(redex_id(x_1)) CHECK_ID(redex_id(x_1)) -> REDUCE(x_1) REDUCE(s(x_1)) -> REDUCE(x_1) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_3) The TRS R consists of the following rules: redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) The set Q consists of the following terms: redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_1) REDUCE(f(x_1, x_2, x_3)) -> CHECK_F(redex_f(x_1, x_2, x_3)) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_2) REDUCE(id(x_1)) -> CHECK_ID(redex_id(x_1)) CHECK_ID(redex_id(x_1)) -> REDUCE(x_1) REDUCE(s(x_1)) -> REDUCE(x_1) CHECK_F(redex_f(x_1, x_2, x_3)) -> REDUCE(x_3) Strictly oriented rules of the TRS R: redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(CHECK_F(x_1)) = 2 + x_1 POL(CHECK_ID(x_1)) = 2 + x_1 POL(REDUCE(x_1)) = 1 + 2*x_1 POL(f(x_1, x_2, x_3)) = 2 + x_1 + 2*x_2 + 2*x_3 POL(id(x_1)) = 2 + x_1 POL(redex_f(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3 POL(redex_id(x_1)) = 2*x_1 POL(result_f(x_1)) = 2 + x_1 POL(result_id(x_1)) = x_1 POL(s(x_1)) = 2 + x_1 ---------------------------------------- (13) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) The set Q consists of the following terms: redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_f(result_f(x)) -> go_up(x) check_id(result_id(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_id(result_id(x)) -> go_up(x) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. top(go_up(x0)) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_id(result_id(x)) -> go_up(x) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) The set Q consists of the following terms: reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule TOP(go_up(x)) -> TOP(reduce(x)) at position [0] we obtained the following new rules [LPAR04]: (TOP(go_up(f(x0, x1, x2))) -> TOP(check_f(redex_f(x0, x1, x2))),TOP(go_up(f(x0, x1, x2))) -> TOP(check_f(redex_f(x0, x1, x2)))) (TOP(go_up(id(x0))) -> TOP(check_id(redex_id(x0))),TOP(go_up(id(x0))) -> TOP(check_id(redex_id(x0)))) (TOP(go_up(s(x0))) -> TOP(in_s_1(reduce(x0))),TOP(go_up(s(x0))) -> TOP(in_s_1(reduce(x0)))) ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(f(x0, x1, x2))) -> TOP(check_f(redex_f(x0, x1, x2))) TOP(go_up(id(x0))) -> TOP(check_id(redex_id(x0))) TOP(go_up(s(x0))) -> TOP(in_s_1(reduce(x0))) The TRS R consists of the following rules: reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_id(result_id(x)) -> go_up(x) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) The set Q consists of the following terms: reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_id(result_id(x)) -> go_up(x) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. top(go_up(x0)) ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(f(x_1, x_2, x_3)) -> check_f(redex_f(x_1, x_2, x_3)) reduce(id(x_1)) -> check_id(redex_id(x_1)) reduce(s(x_1)) -> in_s_1(reduce(x_1)) in_s_1(go_up(x_1)) -> go_up(s(x_1)) redex_id(s(x)) -> result_id(s(id(x))) redex_id(0) -> result_id(0) check_id(result_id(x)) -> go_up(x) check_id(redex_id(x_1)) -> in_id_1(reduce(x_1)) in_id_1(go_up(x_1)) -> go_up(id(x_1)) redex_f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> result_f(f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2, x_3)) -> in_f_1(reduce(x_1), x_2, x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_2(x_1, reduce(x_2), x_3) check_f(redex_f(x_1, x_2, x_3)) -> in_f_3(x_1, x_2, reduce(x_3)) in_f_3(x_1, x_2, go_up(x_3)) -> go_up(f(x_1, x_2, x_3)) in_f_2(x_1, go_up(x_2), x_3) -> go_up(f(x_1, x_2, x_3)) in_f_1(go_up(x_1), x_2, x_3) -> go_up(f(x_1, x_2, x_3)) The set Q consists of the following terms: reduce(f(x0, x1, x2)) reduce(id(x0)) redex_f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) redex_id(s(x0)) redex_id(0) check_f(result_f(x0)) check_id(result_id(x0)) check_f(redex_f(x0, x1, x2)) check_id(redex_id(x0)) reduce(s(x0)) in_f_1(go_up(x0), x1, x2) in_f_2(x0, go_up(x1), x2) in_f_3(x0, x1, go_up(x2)) in_s_1(go_up(x0)) in_id_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) Trivial-Transformation (SOUND) We applied the Trivial transformation to transform the outermost TRS to a standard TRS. ---------------------------------------- (28) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 Q is empty. ---------------------------------------- (29) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is id(s(x)) -> s(id(x)) id(0) -> 0 The TRS R 2 is f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) The signature Sigma is {f_3} ---------------------------------------- (30) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) ---------------------------------------- (31) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> ID(s(s(s(s(s(s(s(s(x))))))))) ID(s(x)) -> ID(x) The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (34) Complex Obligation (AND) ---------------------------------------- (35) Obligation: Q DP problem: The TRS P consists of the following rules: ID(s(x)) -> ID(x) The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (37) Obligation: Q DP problem: The TRS P consists of the following rules: ID(s(x)) -> ID(x) R is empty. The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (38) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) ---------------------------------------- (39) Obligation: Q DP problem: The TRS P consists of the following rules: ID(s(x)) -> ID(x) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (40) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *ID(s(x)) -> ID(x) The graph contains the following edges 1 > 1 ---------------------------------------- (41) YES ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: f(s(s(s(s(s(s(s(s(x)))))))), y, y) -> f(id(s(s(s(s(s(s(s(s(x))))))))), y, y) id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1) ---------------------------------------- (46) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) at position [0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(id(s(s(s(s(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(id(s(s(s(s(s(s(s(x))))))))), y, y)) ---------------------------------------- (48) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(id(s(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (49) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(id(s(s(s(s(s(s(s(x))))))))), y, y) at position [0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(id(s(s(s(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(id(s(s(s(s(s(s(x))))))))), y, y)) ---------------------------------------- (50) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(id(s(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (51) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(id(s(s(s(s(s(s(x))))))))), y, y) at position [0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(id(s(s(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(id(s(s(s(s(s(x))))))))), y, y)) ---------------------------------------- (52) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(id(s(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (53) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(id(s(s(s(s(s(x))))))))), y, y) at position [0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(id(s(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(id(s(s(s(s(x))))))))), y, y)) ---------------------------------------- (54) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(id(s(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (55) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(id(s(s(s(s(x))))))))), y, y) at position [0,0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(id(s(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(id(s(s(s(x))))))))), y, y)) ---------------------------------------- (56) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(id(s(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (57) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(id(s(s(s(x))))))))), y, y) at position [0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(id(s(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(id(s(s(x))))))))), y, y)) ---------------------------------------- (58) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(id(s(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (59) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(id(s(s(x))))))))), y, y) at position [0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(id(s(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(id(s(x))))))))), y, y)) ---------------------------------------- (60) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(id(s(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (61) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(id(s(x))))))))), y, y) at position [0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(s(id(x))))))))), y, y),F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)) ---------------------------------------- (62) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(s(id(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 The set Q consists of the following terms: id(s(x0)) id(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (63) MNOCProof (EQUIVALENT) We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set. ---------------------------------------- (64) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(s(s(s(s(s(s(s(x)))))))), y, y) -> F(s(s(s(s(s(s(s(s(id(x))))))))), y, y) The TRS R consists of the following rules: id(s(x)) -> s(id(x)) id(0) -> 0 Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (65) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = F(s(s(s(s(s(s(s(s(x)))))))), y, y) evaluates to t =F(s(s(s(s(s(s(s(s(id(x))))))))), y, y) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [x / id(x)] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from F(s(s(s(s(s(s(s(s(x)))))))), y, y) to F(s(s(s(s(s(s(s(s(id(x))))))))), y, y). ---------------------------------------- (66) NO